Print indices of array elements whose removal makes the sum of odd and even-indexed elements equal
Given an array arr[] of size N, the task is to find indices of array elements whose removal makes the sum of odd and even indexed elements equal. If no such element exists, then print -1.
Examples:
Input: arr[] = {4, 1, 6, 2}
Output: 1
Explanation:
Removing arr[1] modifies arr[] to {4, 6, 2}
Sum of even-indexed array elements = arr[0] + arr[2] = 4 + 2 = 6
Sum of odd-indexed array elements = arr[1] = 6
Therefore, the required output is 1.Input:arr = {3, 2, 1}
Output: -1
Naive Approach: The simplest approach to solve this problem to traverse the array and remove the ith element from the array and check if the sum of odd-indexed array elements equal to the sum of even-indexed array elements or not. If found to be true then print the index.
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach: The above approach can be optimized using the Prefix Sum technique. The idea is to use the fact that If an array element removed from the index, then after that index, the even-indexed array elements become the odd-indexed and vice-versa. Follow the steps below to solve the problem:
- Initialize two arrays, say odd[] and even[], to store the prefix sum of odd and even-indexed array elements and prefix sum of even-indexed array elements respectively.
- Traverse the array arr[] and compute prefix sum of odd and even-indexed array elements.
- Initialize two variables, say P = 0 and Q = 0, to store the sum of even and odd-indexed array elements respectively after removal of an array element.
- Traverse the array and remove array elements one by one and update prefix sums accordingly. Check if the sum of even-indexed array elements, P, is equal to the sum of odd-indexed array elements, Q or not. If found to be true, then print the current index.
- Otherwise, print -1.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to find indices of array elements // whose removal makes the sum of odd and // even indexed array elements equal void removeIndicesToMakeSumEqual(vector< int >& arr) { // Stores size of array int N = arr.size(); // Store prefix sum of odd // index array elements vector< int > odd(N, 0); // Store prefix sum of even // index array elements vector< int > even(N, 0); // Update even[0] even[0] = arr[0]; // Traverse the given array for ( int i = 1; i < N; i++) { // Update odd[i] odd[i] = odd[i - 1]; // Update even[i] even[i] = even[i - 1]; // If the current index // is an even number if (i % 2 == 0) { // Update even[i] even[i] += arr[i]; } // If the current index // is an odd number else { // Update odd[i] odd[i] += arr[i]; } } // Check if at least one // index found or not that // satisfies the condition bool find = 0; // Store odd indices sum by // removing 0-th index int p = odd[N - 1]; // Store even indices sum by // removing 0-th index int q = even[N - 1] - arr[0]; // If p and q are equal if (p == q) { cout << "0 " ; find = 1; } // Traverse the array arr[] for ( int i = 1; i < N; i++) { // If i is an even number if (i % 2 == 0) { // Update p by removing // the i-th element p = even[N - 1] - even[i - 1] - arr[i] + odd[i - 1]; // Update q by removing // the i-th element q = odd[N - 1] - odd[i - 1] + even[i - 1]; } else { // Update q by removing // the i-th element q = odd[N - 1] - odd[i - 1] - arr[i] + even[i - 1]; // Update p by removing // the i-th element p = even[N - 1] - even[i - 1] + odd[i - 1]; } // If odd index values sum is equal // to even index values sum if (p == q) { // Set the find variable find = 1; // Print the current index cout << i << " " ; } } // If no index found if (!find) { // Print not possible cout << -1; } } // Driver Code int main() { vector< int > arr = { 4, 1, 6, 2 }; removeIndicesToMakeSumEqual(arr); return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GFG{ // Function to find indices of array elements // whose removal makes the sum of odd and // even indexed array elements equal static void removeIndicesToMakeSumEqual( int []arr) { // Stores size of array int N = arr.length; // Store prefix sum of odd // index array elements int []odd = new int [N]; // Store prefix sum of even // index array elements int []even = new int [N]; // Update even[0] even[ 0 ] = arr[ 0 ]; // Traverse the given array for ( int i = 1 ; i < N; i++) { // Update odd[i] odd[i] = odd[i - 1 ]; // Update even[i] even[i] = even[i - 1 ]; // If the current index // is an even number if (i % 2 == 0 ) { // Update even[i] even[i] += arr[i]; } // If the current index // is an odd number else { // Update odd[i] odd[i] += arr[i]; } } // Check if at least one // index found or not that // satisfies the condition boolean find = false ; // Store odd indices sum by // removing 0-th index int p = odd[N - 1 ]; // Store even indices sum by // removing 0-th index int q = even[N - 1 ] - arr[ 0 ]; // If p and q are equal if (p == q) { System.out.print( "0 " ); find = true ; } // Traverse the array arr[] for ( int i = 1 ; i < N; i++) { // If i is an even number if (i % 2 == 0 ) { // Update p by removing // the i-th element p = even[N - 1 ] - even[i - 1 ] - arr[i] + odd[i - 1 ]; // Update q by removing // the i-th element q = odd[N - 1 ] - odd[i - 1 ] + even[i - 1 ]; } else { // Update q by removing // the i-th element q = odd[N - 1 ] - odd[i - 1 ] - arr[i] + even[i - 1 ]; // Update p by removing // the i-th element p = even[N - 1 ] - even[i - 1 ] + odd[i - 1 ]; } // If odd index values sum is equal // to even index values sum if (p == q) { // Set the find variable find = true ; // Print the current index System.out.print(i + " " ); } } // If no index found if (!find) { // Print not possible System.out.print(- 1 ); } } // Driver Code public static void main(String[] args) { int [] arr = { 4 , 1 , 6 , 2 }; removeIndicesToMakeSumEqual(arr); } } // This code is contributed by Amit Katiyar |
Python3
# Python program to implement # the above approach # Function to find indices of array elements # whose removal makes the sum of odd and # even indexed array elements equal def removeIndicesToMakeSumEqual(arr): # Stores size of array N = len (arr); # Store prefix sum of odd # index array elements odd = [ 0 ] * N; # Store prefix sum of even # index array elements even = [ 0 ] * N; # Update even[0] even[ 0 ] = arr[ 0 ]; # Traverse the given array for i in range ( 1 , N): # Update odd[i] odd[i] = odd[i - 1 ]; # Update even[i] even[i] = even[i - 1 ]; # If the current index # is an even number if (i % 2 = = 0 ): # Update even[i] even[i] + = arr[i]; # If the current index # is an odd number else : # Update odd[i] odd[i] + = arr[i]; # Check if at least one # index found or not that # satisfies the condition find = False ; # Store odd indices sum by # removing 0-th index p = odd[N - 1 ]; # Store even indices sum by # removing 0-th index q = even[N - 1 ] - arr[ 0 ]; # If p and q are equal if (p = = q): print ( "0 " ); find = True ; # Traverse the array arr for i in range ( 1 , N): # If i is an even number if (i % 2 = = 0 ): # Update p by removing # the i-th element p = even[N - 1 ] - even[i - 1 ] - arr[i] + odd[i - 1 ]; # Update q by removing # the i-th element q = odd[N - 1 ] - odd[i - 1 ] + even[i - 1 ]; else : # Update q by removing # the i-th element q = odd[N - 1 ] - odd[i - 1 ] - arr[i] + even[i - 1 ]; # Update p by removing # the i-th element p = even[N - 1 ] - even[i - 1 ] + odd[i - 1 ]; # If odd index values sum is equal # to even index values sum if (p = = q): # Set the find variable find = True ; # Print the current index print (i, end = ""); # If no index found if (find = = False ): # Print not possible print ( - 1 ); # Driver Code if __name__ = = '__main__' : arr = [ 4 , 1 , 6 , 2 ]; removeIndicesToMakeSumEqual(arr); # This code is contributed by shikhasingrajput |
C#
// C# program to implement // the above approach using System; class GFG{ // Function to find indices of array elements // whose removal makes the sum of odd and // even indexed array elements equal static void removeIndicesToMakeSumEqual( int []arr) { // Stores size of array int N = arr.Length; // Store prefix sum of odd // index array elements int []odd = new int [N]; // Store prefix sum of even // index array elements int []even = new int [N]; // Update even[0] even[0] = arr[0]; // Traverse the given array for ( int i = 1; i < N; i++) { // Update odd[i] odd[i] = odd[i - 1]; // Update even[i] even[i] = even[i - 1]; // If the current index // is an even number if (i % 2 == 0) { // Update even[i] even[i] += arr[i]; } // If the current index // is an odd number else { // Update odd[i] odd[i] += arr[i]; } } // Check if at least one // index found or not that // satisfies the condition bool find = false ; // Store odd indices sum by // removing 0-th index int p = odd[N - 1]; // Store even indices sum by // removing 0-th index int q = even[N - 1] - arr[0]; // If p and q are equal if (p == q) { Console.Write( "0 " ); find = true ; } // Traverse the array arr[] for ( int i = 1; i < N; i++) { // If i is an even number if (i % 2 == 0) { // Update p by removing // the i-th element p = even[N - 1] - even[i - 1] - arr[i] + odd[i - 1]; // Update q by removing // the i-th element q = odd[N - 1] - odd[i - 1] + even[i - 1]; } else { // Update q by removing // the i-th element q = odd[N - 1] - odd[i - 1] - arr[i] + even[i - 1]; // Update p by removing // the i-th element p = even[N - 1] - even[i - 1] + odd[i - 1]; } // If odd index values sum is equal // to even index values sum if (p == q) { // Set the find variable find = true ; // Print the current index Console.Write(i + " " ); } } // If no index found if (!find) { // Print not possible Console.Write(-1); } } // Driver Code public static void Main(String[] args) { int [] arr = { 4, 1, 6, 2 }; removeIndicesToMakeSumEqual(arr); } } // This code is contributed by AnkThon |
Javascript
<script> // javascript program to implement // the above approach // Function to find indices of array elements // whose removal makes the sum of odd and // even indexed array elements equal function removeIndicesToMakeSumEqual(arr) { // Stores size of array var N = arr.length; // Store prefix sum of odd // index array elements var odd = Array(N).fill(0); // Store prefix sum of even // index array elements var even = Array(N).fill(0); // Update even[0] even[0] = arr[0]; // Traverse the given array for (i = 1; i < N; i++) { // Update odd[i] odd[i] = odd[i - 1]; // Update even[i] even[i] = even[i - 1]; // If the current index // is an even number if (i % 2 == 0) { // Update even[i] even[i] += arr[i]; } // If the current index // is an odd number else { // Update odd[i] odd[i] += arr[i]; } } // Check if at least one // index found or not that // satisfies the condition var find = false ; // Store odd indices sum by // removing 0-th index var p = odd[N - 1]; // Store even indices sum by // removing 0-th index var q = even[N - 1] - arr[0]; // If p and q are equal if (p == q) { document.write( "0 " ); find = true ; } // Traverse the array arr for (i = 1; i < N; i++) { // If i is an even number if (i % 2 == 0) { // Update p by removing // the i-th element p = even[N - 1] - even[i - 1] - arr[i] + odd[i - 1]; // Update q by removing // the i-th element q = odd[N - 1] - odd[i - 1] + even[i - 1]; } else { // Update q by removing // the i-th element q = odd[N - 1] - odd[i - 1] - arr[i] + even[i - 1]; // Update p by removing // the i-th element p = even[N - 1] - even[i - 1] + odd[i - 1]; } // If odd index values sum is equal // to even index values sum if (p == q) { // Set the find variable find = true ; // Print the current index document.write(i + " " ); } } // If no index found if (!find) { // Print not possible document.write(-1); } } // Driver Code var arr = [ 4, 1, 6, 2 ]; removeIndicesToMakeSumEqual(arr); // This code is contributed by Rajput-Ji </script> |
1
Time Complexity: O(N)
Auxiliary Space: O(N)
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